2.2.24 | revision #12583 | Savchenko Solutions

Statement

2.2.24^∗. Two bodies of mass $$m_1$$ and $$m_2$$ are connected by a stretched thread of length $$l$$ and move along a smooth horizontal surface. At some point in time, it turned out that the first body is stationary, and the velocity of the second body, equal to $$v$$ , is perpendicular to the thread. Determine the tension force of the thread.

Solution

Since point $$1$$ is at rest, it follows that the forces acting on it are compensated

$\vec{F}_{c1}=-\vec{T}_{1}$

$T_1=m_1\omega^2 x$

Where distance $$x$$ between point $$1$$ and center of mass

$x=l\frac{m_2}{m_1+m_2}$

Direction of forces and velocity of the centre of mass

Since point $$1$$ is at rest, the motion is around it. Then the angular velocity of rotation is found through the velocity $$v$$ of the point $$2$$

$\omega = \frac{v}{l}$

Now, substitute all of this into the expression for $$T_1$$

$T_1=m_1\frac{v^2}{l^2} \cdot l\frac{m_2}{m_1+m_2}$

$T_1=\frac{m_1m_2v^2}{(m_1+m_2)l}$

According to Newton's third law, since the thread is weightless, the absolute value of tension force of the thread at point $$1$$ and point $$2$$ are equal.

$$$T_1=T_2=T$$$

$\boxed{T=\frac{m_1m_2}{m_1+m_2}\frac{v^2}{l}}$

Answer

$F=\frac{m_1m_2v^2}{(m_1+m_2)l}$

Alternative solution

Let's consider this system as different bodies. Using Newton's Second law of motion, we can get:

$\begin{cases} \vec{F}_1 = m_1 \vec{a}_1 \\ \vec{F}_2 = m_2 \vec{a}_2 \end{cases} \Rightarrow \begin{cases} \vec{a}_1 = \frac{\vec{F}_1}{m_1} \\ \vec{a}_2 = \frac{\vec{F}_2}{m_2} \end{cases} \quad \text{(1)}$

Let's subtract $\vec{a}_2$ from $\vec{a}_1$ :

$\vec{a}_1 - \vec{a}_2 = \frac{\vec{F}_1}{m_1} - \frac{\vec{F}_2}{m_2} \quad \text{(2)}$

As we can see, it looks like the derivative of relative velocity:

$\vec{v}_{B/A} = \vec{v}_B - \vec{v}_A \quad \text{(3)}$

Now, let's solve the derivative:

$\frac{d(\vec{v}_{B/A})}{dt} = \frac{d(\vec{v}_B)}{dt} - \frac{d(\vec{v}_A)}{dt} \Rightarrow \vec{a}_{B/A} = \vec{a}_B - \vec{a}_A \quad \text{(4)}$

This is why we can say that:

$\vec{a}_1 - \vec{a}_2 = \vec{a}_{12} = \frac{\vec{F}_1}{m_1} - \frac{\vec{F}_2}{m_2} \quad \text{(5)}$

Due to the weightlessness of the thread and Newton's Third Law of Motion $\vec{F}_1 = -\vec{F}_2 = \vec{F}$ : Eventually:

$\vec{a}_{12} = \vec{F} \left( \frac{1}{m_1} + \frac{1}{m_2} \right) = \vec{F} \frac{m_1 + m_2}{m_1 m_2} \quad \text{(6)}$

$a_{12} = \frac{v^2}{l}$ , where $$l$$ is the length of the thread, and $$v$$ is the relative velocity

$F = \frac{m_1 m_2}{m_1 + m_2} \frac{v^2}{l} \quad \text{(7)}$